Integral Dupla Iterada UNIP

1)Calcule o volume dos sólidos

a)f(x,y)=6-2x-2y

∫_0^1▒〖∫_0^2▒〖6-2x-2y dydx〗=[6y-2xy-y^2]〗 2¦0 0≤x≤1;0≤y≤2

=(6(2)-2x(2)-(2)^2 )=12-4x-4

=8-4x

∫_0^1▒〖8-4x dx=[8x-2x^2]1¦0〗

=(8(1)-2(1)^2 )=8-2

=6

b)f(x,y)=1/xy=1/x.1/y ∫_1^2▒∫_1^3▒〖1/x.1/y 〗 dydx =[1/x.ln⁡〖(y)〗 ]_1^3

( 1/x.ln⁡(3) )-( 1/x.ln⁡(1) ) = ( 1/x.ln⁡(3) )-0 =( 1/x.ln⁡(3) )

∫_1^2▒( 1/x.ln⁡(3) )dx=[ln⁡(x).(ln3)] ■(2@1) =(ln⁡(2).(ln3))-(ln⁡(1)-ln⁡(3) ) =(ln⁡(2).(ln3))-0=ln⁡(2).ln⁡(3)

c)f(x,y)=e^(x+y) ∫_0^1▒∫_0^ln⁡(2)▒〖e^x.e^y 〗 =∫_0^ln⁡(2)▒〖e^x.e^y 〗 =[e^x.e^y ] ln⁡(2)¦0

=(e^x.e^ln⁡(2) )-(e^x.e^0 ) =(e^x.2)-(e^x.1) =2e^x-e^x

∫_0^1▒〖2e^x-e^x dx〗 =[2e^x-e^x ] 1¦0 =(2e^1-e^1 )-(2e^0-e^0 )

e^0=1

=(2e-e)-(2-1) =1e-1 =e-1

d)f(x,y)=(1-x).(4-y)

∫_0^1▒∫_0^4▒〖(1-x).(4-y) 〗 dydx =∫_0^4▒(4-y-4x+xy)dy =[4y-y^2-4xy+(xy^2)/2]_0^4

(4(4)-(4)^2/2-4x(4)+(x(4)^2)/2) =(16-8-16x+8x) =8-8x

∫_0^1▒〖8-8x dx〗=[8x-(8x^2)/2]_0^1 =[8x-4x^2 ] 1¦0 =(8(1)-(4(1)^2 )

=8-4 =4

2.Calcule as integrais iteradas.

a)∫_0^1▒〖∫_0^(x^2)▒〖x+2y dydx〗= [xy+y^2 ] x^2¦0〗 =(x(x^2 )+(x^2 )^2 ) =x^3+x^4

∫_0^1▒〖x^3+x^4 dx〗 =[x^4/4+x^5/5]_0^1 =(1^4/4+1^5/5) =1/4+1/5 =(5+4)/20 =9/20

b)∫_1^2▒〖∫_y^2▒〖xy 〗 〗 dxdy =[x^2/2.y]_y^2 =((2)^2/2 y)-((y)^2/2 y) =2y-y^3/2

∫_1^2▒〖2y-y^3/2〗 dy =[y^2-y^4/8]_1^2 =((2)^2-2^4/8)-(1^2-1^4/8) =(4-16/8)-(1-1/8) =((32-16)/8)-((8-1)/1) =(16/8-7/8) =(16-7)/8 =9/8

c)∫_0^1▒〖∫_x^(2-x)▒〖x^2-y〗 〗 dydx =[x^2 y-y^2/2]_x^(2-x) =(x^2 (2-x)-(2-x)^2/2)-(x^2 x-x^2/2) =(2x^2-x^3-2+2x-x^2/2)-x^3+x^2/2 =2x^2-x^3-2+2x-x^2/2-x^3+x^2/2 =2x^2-2x^3-2+2x

∫_0^1▒〖2x^2-2x^3-2+2x〗 dx =[(2x^3)/3-(2x^4)/4-2x+(2x^2)/2]_0^1 =[(2x^3)/3-x^4/2-2x+x^2 ]_0^1 =(2/3-1/2-2+1) =(4-3-12+6)/6 =-5/6

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